# 1 = 2

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Many people have seen “proofs” that 1 = 2. However most of them have an obvious fallacy – a division by 0.

A classic such example is:

\begin{eqnarray*}
a &=& b, \\
a^2 &=& ab, \\
a^2 – b^2 &=& ab – b^2, \\
(a+b)(a-b) &=& b(a-b), \\
(a+b) &=& b, \\
a+a &=& a, \\
2a &=& a, \\
2 &=& 1.
\end{eqnarray*}

This is a decent “trick” to show people, but as I said above, the mistake simply comes from the division by 0 on line 5. However, here is a different “proof”, can you see the mistake?

Consider the equation, $$2=x^{x^{x^{\ldots}}}$$, with an infinite number of $$x$$’s. If we add some brackets to the equation we get, $$2=x^{\left(x^{x^{\ldots}}\right)}$$ and then we can substitute the first equation in to get: $2=x^2.$ So $$x=\sqrt{2}$$, and hence $2=\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\ldots}}}$.

Then if we repeat the exact same above process with the equation $$4=x^{x^{x^{\ldots}}}$$ we see that $$4=x^4$$ and so again $$x=\sqrt{2}$$. So this time, $4=\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\ldots}}}.$

And so $$2=4$$ and thus $$1=2$$!

Where is the mistake? ## One comment

1. alex says:

Is it real or just for fun?

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